# Three phase system - PowerPoint PPT Presentation

PPT – Three phase system PowerPoint presentation | free to download - id: 3e14e3-OTAzN

The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
Title:

## Three phase system

Description:

### Three phase system Three-wire star-connected system with balanced load Unbalanced load Power in three phase Three phase system Three-wire star-connected system with ... – PowerPoint PPT presentation

Number of Views:129
Avg rating:3.0/5.0
Slides: 34
Provided by: PROF104
Category:
Tags:
Transcript and Presenter's Notes

Title: Three phase system

1
Three phase system
2
Single phase generator
Current induces in the coil as the coil moves in
the magnetic field
Generator for single phase
Note Induction motor cannot start by itself. This
problem is solved by introducing three phase
system
Current produced at terminal
3
Three phase generator
Instead of using one coil only , three coils are
used arranged in one axis with orientation of
120o each other. The coils are R-R1 , Y-Y1 and
B-B1. The phases are measured in this sequence
R-Y-B. I.e Y lags R by 120o , B lags Y by 120o.
4
The three winding can be represented by the above
circuit. In this case we have six wires. The emf
are represented by eR , eY, eB.
5
The circuit can be simplified as follows, where
R1 can be connected to Y and Y1 can be connected
to B. In this case the circuit is reduced to 4
wires.
Since the total emf is zero, R and B1 can be
connected together, thus we arrive with delta
connection system.
6
Delta connection of three phase windings
Fig. B
Fig.A
Fig. C
• Since the total emf is zero, R and B1 can be
connected together as in Fig.A , thus we arrive
with delta connection system as in Fig. C.
• The direction of the emf can be referred to the
emf waveform as in Fig. B where PL is ve (R1-R),
PM is ve (Y-Y1) and PN is ve (B-B1).

7
Star connection of three phase windings
• R1, Y1 and B1 are connected together.
• As the e.m.f generated are assumed in positive
direction , therefore the current directions are
also considered as flowing in the positive
direction.
• The current in the common wire (MN) is equal to
the sum of the generated currents. i.e iRiYiB
.
• This arrangement is called four wire
star-connected system. The point N refers to star
point or neutral point.

8
The instantaneous current in loads L1 , L2 and L3
are
9
Three-wire star-connected system with balanced
For balanced loads, the fourth wire carries no
current , so it can be dispensed
10
Instantaneous currents waveform for iR, iY and
iB in a balanced three-phase system.
11
Voltage and current in star connection
• VRY, VYB and VBR are called line voltage
• VR, VY and VB are called phase voltage

From Kirchoff voltage law we have
In phasor diagram
12
For balanced load VR , VY and VB are equaled but
out of phase
VRY VL?30? VYB VL?-90? VBR VL?150?
VR VP?30? VY VP?-90? VB VP?150?
therefore
13
then
and
14
Voltage and current in Delta connection
• IR, IY and IB are called line current
• I1, I2 and I3 are called phase current

From Kirchoff current law we have
VP
VL
In phasor diagram
15
Since the loads are balanced, the magnitude of
currents are equaled but 120o out of phase. i.e
I1 I2I3 ,IP Therefore-
I1 VP?30? I2 VP?-90? I3 VP?150?
IR IL?30? IY IL?-90? IB IL?150?
Where IP is a phase current and IL is a line
current
Thus IRIYIB IL
16
Hence
17
Example 1
In a three-phase four-wire system the line
voltage is 400V and non-inductive loads of 5 kW,
8 kW and 10 kW are connected between the three
conductors and the neutral. Calculate (a) the
current in each phase (b) the current in the
neutral conductor.
18
Voltage to neutral
Current in 10kW resistor
Current in 8kW resistor
Current in 5kW resistor
19
Resolve the current components into horizontal
and vertical components.
20
Example 2
• A delta connected load is arranged as in Figure
below. The supply voltage is 400V at 50Hz.
Calculate
• The phase currents
• The line currents.

(a)
I1 is in phase with VRY since there is only
resistor in the branch
21
In branch between YB , there are two components ,
R2 and X2
In the branch RB , only capacitor in it , so the
XC is -90 out of phase.
22
(b)
q30o
q 71o 34 -60o 11o 34
23
q 180-30o-11o 34 138o 34
24
Power in three phase
Active power per phase IPVP x power factor
Total active power 3VPIP x power factor
If IL and VL are rms values for line current and
line voltage respectively. Then for delta (?)
connection VP VL and IP IL/?3. therefore
For star connection (?) VP VL/?3 and IP
IL. therefore
25
Example 3
• A three-phase motor operating off a 400V system
is developing 20kW at an efficiency of 0.87 p.u
and a power factor of 0.82. Calculate
• The line current
• The phase current if the windings are
delta-connected.

(a) Since
And line current IL40.0A
(b) For a delta-connected winding
26
Example 4
Three identical coils, each having a resistance
of 20? and an inductance of 0.5 H connected in
(a) star and (b) delta to a three phase supply of
400 V 50 Hz. Calculate the current and the
total power absorbed by both method of
connections.
First of all calculating the impedance of the
coils
where
27
Star-connection
Since it is a balanced load
Power absorbed
28
Star connection
29
Example 5
A balanced three phase load connected in star,
each phase consists of resistance of 100 ?
paralleled with a capacitance of 31.8 ?F. The
load is connected to a three phase supply of 415
V 50 Hz. Calculate (a) the line
current (b) the power absorbed (c) total
kVA (d) power factor .
30
where
Line current
Volt-ampere per phase
Active power per phase
Total active power
31
(b)
Reactive power per phase
Total reactive power
Total volt-ampere
(c)
(d)
Power Factor cos? cos 45? 0.707 (leading)
32
Example 6
A three phase star-connected system having a
phase voltage of 230V and loads consist of non
reactive resistance of 4 ?, 5 ? and 6?
respectively. Calculate (a) the current in each
phase conductor (b) the current in neutral
conductor and (c) total power absorbed.
33
(b)
X-component 46 cos 30? 38.3 cos 30? - 57.5
15.5 A Y-component 46 sin 30? - 38.3 sin 30?
3.9 A
Therefore
(c)